# Vout of Common Emitter Amplifier

Discussion in 'Homework Help' started by student_technician, Dec 28, 2014.

1. ### student_technician Thread Starter New Member

Dec 28, 2014
18
0
I have been trying to solve the Vout of a common emitter amplifier....This is the last step in a full analysis of the amplifier. It was my understanding that I could use Av and multiply by the Vin (AC) to get the Output voltage. When I do this I get a different answer then the key provided my Instructors.

Vin=500mVpp
Rs=270 ohms
Rintot=4.2k ohms
Av=18.2
Rout=2k ohms
RL=5k ohms
Vout=? (ans=6.11Vpp)

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
Where is Rs located?
What is Rintot and where is it located?
Is Rout located between collector and ground?
Where is RL located?

"Picture is worth a thousand words."

3. ### student_technician Thread Starter New Member

Dec 28, 2014
18
0
Here is a pic of amplifier and model

Rintotal=R1//R2//Rinbase which is labeled as Rin in the lower model

I just noticed I made an error in my sketch: the 6.8k Resistor should be grounded instead of connected to the the same node as the cap.

Last edited: Dec 28, 2014
4. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,500
511
Ok. I got 6.107 volts peak-to-peak, so round it up and you get 6.11 volts peak to peak.

Here is what is going on. The 500 mVpp is going into voltage divider. Do you see it? The voltage divider is formed by 270 Ohm and 4.2 kOhm resistors. Notice that voltage across 4.2 kOhm resistor is the voltage being seen by the amplifier. So Step 1, find voltage across 4.2 kOhm resistor (there is a standard formula for it).

Step 2. Multiply the voltage across 4.2 kOhm (Vin) by the gain (Av).

Now Step 3. The output of the amplifier is input to another voltage divider. Do you see it? I assume the value you are looking for is the voltage across RL. So, you know the voltage that is going into the voltage divider formed by 2 kOhm and 5 kOhm resistors, you know the values of the resistors that form the voltage divider, all you have to do is solve for the voltage across the 5 kOhm resistor.

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5. ### student_technician Thread Starter New Member

Dec 28, 2014
18
0
Thank you for the help, that makes great sense I just don't remember going over that in class (not in my notes either).